The relative decreases in the vapour pressure of an aqueous solution containing `2 mol [Cu(NH_(3)_(3)Cl]` in `3 mol H_(2)O` is `0.50`. On reaction with `AgNO_(3)`, this solution will form

Options

A. `1 mol AgCl`
B. `0.25 mol AgCl`
C. `2 mol AgCl`
D. `0.40 mol AgCl`

Answer

A

Explanation

`(DeltaP)/P^(@)=0.50 = (n_(1)i)/(n_(1)i + n_(2))`
`0.50 =(2i)/(2i + 3)`
`i=1.5 = 1 + x`
`x=0.5`
Hence,2 mol NaCl will exists as `1 "mol" Cl^(ө)` due to `50%` ionization.